For the second statement in the theorem, we consider the remaining two cases that \(\mathbb {E}[X\mid Y]\)  is a linear expression in \(Y\)  or is a constant. If \(\mathbb {E}[X\mid Y]\)  is constant, then differentiating and taking the expectation in (REF ) yields that \(\Vert X-\mathbb {E}[X\mid Y]\Vert _2=0,\)  i.e., \(X=\mathbb {E}[X\mid Y]\)  is constant. Finally, under the assumption that \(X\)  has finite variance, \(\mathbb {E}[X \mid Y]\)  is linear if and only if \(X\)  is Gaussian (see, e.g., [1]}). We note that if one requires only that \(X\)  be integrable, then one may deduce directly from the differential equation (REF ) that a linear \(\mathbb {E}[X\mid Y]\)  implies a Gaussian \(X\)  in this case too, and, for completeness, we end with a proof of this fact.
